A full-scale model. Credit: NASA |

Longer means infrared. Being an infrared telescope, JWST will see through dust, directly image planets, and peer further back in time at objects redshifted out of the visible range. But infrared telescopes come with some complications. On Earth, we don't do a lot of infrared astronomy, partly because the atmosphere absorbs too much of it, but also because stuff too cold to emit visible light (basically everything on Earth) is usually spilling out lots of infrared instead. We can't do IR astronomy on Earth for the same reason we can't do visible astronomy during the day: it's too bright.

That's why JWST will be in space. But even in space, the Earth and sun loom large. Keep the telescope too near the Earth, and the Earth warms it up, generating noise in the cameras. JWST must be kept cold, much colder than the objects it wants to look at. The only way to accomplish that is to put it far away from the Earth and hold up a shield to block the Earth and sun. The trick is that you want to be able to block both bodies at the same time, which wouldn't work if you just flung the satellite into any old orbit. The farther you get from the sun, the longer your year (Kepler's third law says the cube of your semi-major axis is proportional to the square of your year), so the sun and Earth will change relative positions in the sky.

You need to find an orbit that's far away, stable, and lets you block two objects at once--tricky. Arranging three objects in space is known as the three-body problem in celestial mechanics , and it has a long history. When Newton first formulated his laws of motion and gravity, he was able to solve the one- and two-body problems. That is, he could tell you how a tiny, insignificant planet would orbit a gigantic star (the one-body problem) or how two comparable objects would orbit each other (the two-body problem), but he was not able to count any higher than 2. Newton reasoned that miniscule interactions from nearby planets would build up over time and slowly destabilize orbits, and he assumed the only solution was divine intervention.

Astronomers, physicists, and mathematicians spent a long time looking for more precise answers. It turns out there is no generic solution to the three-body problem, no simple orbit that works for any configuration of three or more masses. Using perturbation theory, you can account for the infinitesimal, cumulative influences of many bodies over time, but in the long run (millions of years), orbits become chaotic. Chaotic doesn't necessarily mean that a planet will be flung from the solar system, but that we eventually can't say with any precision where in an orbit a planet will be at any given time.

A couple mathematicians were able to work out very specific periodic solutions to what gets called the restricted three-body problem, or the 2+1 body problem: two large gravitating masses, one tiny mass that is virtually insignificant. In just the right location relative to the big ones, the small one can be stable. Nowadays these are known as the Lagrange points, in honor of one of the mathematicians who worked them out (Euler already had enough named after him).

This seems perfect for JWST. If there's a line between the sun and the Earth, we want JWST to be on that line out past Earth. Can we find a Lagrange point there?

In space, lines are purple. |

Notice I said angular speed, which is how long it takes to move a given angle rather than a given distance. If you think about a spinning tire, the outer bits are moving faster than the inner bits, because the bigger the radius, the larger the circumference covered in the same amount of time. But they are both covering the same fraction of a circle in the same time, and thus both have the same angular speed. If different bits moved at different angular speeds, they wouldn’t keep the same relative positions and the tire would spin apart.

We want our frame and JWST to be moving at the same angular speed as the Earth. But in establishing this frame of reference, we have invalidated Newton's laws of motion. We are no longer in an inertial frame, which is one moving at a constant velocity. Circular motion is not constant, because velocity includes direction.

What does it mean for Newton's laws to be invalidated? It means that an object not experiencing any net force will seem to accelerate away. For our rotating frame, maintaining circular motion requires constant force toward the center of the circle. Tie a ball to the end of a string and spin the ball in a circle. The tension along the string is the radial force that maintains circular motion. If the ball comes loose, it will fly off in a straight line. But from the frame of the spinning string, which can continue spinning as long as you supply a force, the ball will appear to curve away. This tendency to accelerate away from a spinning frame can be accounted for if we invent a fictitious force--centrifugal force--that acts in opposition to whatever force maintains circular motion--centripetal force.

So if we look at the Earth from a rotating frame, JWST will seem to experience a centrifugal force pushing it away from the Earth. In order to have the telescope remain stationary in our rotating frame, the force from gravity must balance the centrifugal force.

Doing physics really involves making diagrams like this. |

F

_{g}= Gm

_{sun}m

_{jwst}/(a+d)

^{2}+ Gm

_{earth}m

_{jwst}/d

^{2}

And it's pulled outward by the centrifugal force which results from the angular motion of the system. How do we characterize the centrifugal force? It's the square of the angular speed times the distance from the center of mass (the sun, in this case) times the mass of the accelerating object. The angular speed is inversely proportional to the period, the Earth’s year. So centrifugal force involves the square of the period. Using Kepler’s relation between period and semi-major axis, we can substitute in that quantity (a in our diagram). Doing some algebra, that gives us a centrifugal force of:

F

_{c}= G(m

_{sun}+m

_{earth})m

_{jwst}(a+d)/a

^{3}

And we want F

_{g}to equal F

_{c}. If we cancel some stuff out, we arrive at the following expression, which is defined purely in terms of the masses and the distances between them:

m

_{sun}/(a+d)

^{2}+ m

_{earth}/d

^{2}= (m

_{sun}+m

_{earth})(a+d)/a

^{3}

We're trying to solve for d, the point at which all these forces cancel out. But there's a problem. If we were to multiply all these terms out (FOIL!), we'd find this was a quintic function, which means there'd be a d

^{5}. And there is no equivalent of the quadratic formula for quintic equations. So we have to make some approximations. We have to assume that the sun is so much bigger than the Earth (true, in this case) that the Earth can be ignored whenever the two terms are added together. And we also assume that d is much smaller than a, which lets us do some mathematical tricks. If you make those approximations, and then do some more algebra, you eventually find that:

d = a(m

_{earth}/3m

_{sun})

^{1/3}

That is the location of the second Lagrange point (and the first one, but on the other side). Plugging in the relevant numbers, d = 1.5 million km, which is curiously 1/100 Earth’s distance from the sun. The sun is a little more than a hundred times wider than the Earth, which means that from L2, the Earth and sun appear just about the same size--more or less the moon. And that means JWST can easily block both of them with the same shield. (The similarity in angular size really is a happy coincidence that has to do with an accidental congruence of densities, radii, and that factor of 3 up there. Try it with any other planet and it doesn't work.)

So there you have it. When the combined gravitational pull of the Earth and sun cancel out the centrifugal force pushing JWST away, the telescope remains stationary with respect to Earth’s motion about the sun. It sits 1.5 million km behind the Earth and completes an orbit in a year despite being farther away from the sun.

But that's not quite the end of the story. It turns out that L1, L2, and L3 (on the other side of the sun from the Earth) are only metastable, which means a slight push sends an object flying off into a new orbit. So we can put satellites there, but they require station keeping to prevent them from falling away. L4 and L5, which form equilateral triangles with the two big masses of the 2+1 problem, are stable. Consequently, we actually find families of asteroids called the Trojans at the Sun-Jupiter L4 and L5 points. Also, I’ve totally neglected the Coriolis effect here, which is another fictitious force that pops up when… oh dear, look at that word count.

"If it were easy, everyone would do it."

ReplyDeleteWhat's that actual distance of the L2 point? 930,000 miles? How much does the Moon going by roughly 680,000 miles in destabilize the L2?

ReplyDeleteWell, there are a couple things to consider here. When I talk about making approximations to find the L2 point, those are essentially mathematical tricks to solve the equation. They produce the correct answer, but get there by a method other than algebraic manipulation (because algebraic manipulation alone can't work here). There are other methods I could have used, such as guessing solutions, or graphing the equation and seeing where its roots are. They would all produce the correct answer.

ReplyDeleteBut... my correct answer is an approximation of the real system, because it makes some assumptions about the system. It assumes the center of mass is at the exact center of the sun, it assumes the Earth's orbit is circular, it assumes no Coriolis effect, etc. Another assumption, because this is the 3-body problem, is that no other gravitating masses are significant.

The average distance of the moon from the Earth is 384,400 km, or 238,900 miles. And the moon's mass is small. This works out to the sun's gravity being about 1,500 times stronger at L2 than the moon's, for example. I'm sure this smudges up the numbers a bit in real life, and I'm sure the rocket scientists who keep an eye on satellites out there have to track all that, but saying L2 is at 1.5 million km (or 930,000 miles) is perfectly accurate (if not precise).

The moon and other bodies aren't the reason why L1, L2, and L3 aren't inherently stable, though. That has to do with the shape of the potential wells that create the Lagrange points. Essentially, being at one of those points is like being in a shallow depression at the top of a hill. As long as nothing knocks you out of your little hole, you're okay. But if a slight breeze gives you any momentum, you go rolling down the hill.

Right. It's not a stable point to begin with, which is why I asked if the Moon's influence was significant.

ReplyDeleteSo you're more-or-less saying not very; probably...